# Why is this algebraic manipulation of quaternions incorrect?

I know that for quaternions, $$i^2=j^2=k^2=ijk=-1$$ I've tried to understand this intuitively as having $i$, $j$ and $k$ represent a rotation about each of three axes. But when I do a bit of manipulation with the quaternions, I ran into a problem. If $$i^2=ijk$$ shouldn't $$i=jk \text{ ?}$$ But because $$i^2=-1$$ $$(jk)^2=-1$$ and $$j^2k^2=-1$$ which is obviously not true: $j^2k^2$ should equal $1$.

Where is my error?

By the way, Wikipedia notes that quaternionic multiplication is non-commutative. Does that have something to do with it?

• as mentioned below, your error is that $(jk)^2 \neq j^2k^2$ but instead $(jk)^2 = (jk)(jk) = jkjk$ – JMoravitz Nov 12 '14 at 2:23

I view quaternions as reading. The order is $\{1, i, j, k\}$ so we read left to right. That is, $ij = k$ since that is left to right whereas $ji = -k$ since this is right to left so it is reversed, $-1$. $$(jk)^2 = jkjk = ii = -1$$ Also note that $$j^2k^2 = jjkk = jik = -kk = 1\neq jkjk = ii = -1$$ Does this help?

• So it hinges on $(jk)^2 \neq j^2k^2$? – HDE 226868 Nov 12 '14 at 2:22
• @HDE226868 you changed the order. $jjkk = jik = -kk = 1$ is not $jkjk$ – dustin Nov 12 '14 at 2:23

There is another way of looking at quaternions - they are identical to matrices of a certain kind. Quaternions of the form $bi + cj + dk + a$ can be represented by matrices of the form $$\begin{bmatrix} a+bi & c+di \\ -c+di & a-bi \end{bmatrix}$$ where the $i$ inside the matrix is $\sqrt{-1}$. The matrix representing the quaternion $i$ is $$\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$$ Write down $jk$ in this form $$jk= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$$ which is clearly $i$, not $-i$.

The rule is do nothing with quaternions that you cannot do with matrices of the form given above and you'll be ok.

Edited to add: If you think of $(jk)^2$ in terms of matrix multiplication, $$(jk)^2=jkjk=ii= \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} =-1$$ But $j^2k^2$ = $jjkk = jik =$

$$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =1$$

• I'm guessing that this goes along with the non-commutability? (i.e. for matrices, it is sometimes the case that $AB \neq BA$) – HDE 226868 Nov 12 '14 at 2:39
• @HDE226868 jk is indeed equal to i, for the reasons given above. – user_of_math Nov 12 '14 at 2:44
• @HDE226868 read the quaternion tag. It will tell you it is noncommutative division algebra. – dustin Nov 12 '14 at 2:59
• @HDE226868 Yeah, quaternion algebra is associative but not commutative. – user_of_math Nov 12 '14 at 3:01
• @dustin Yes, I knew that, but wasn't sure how it applied here. – HDE 226868 Nov 12 '14 at 3:09